26.7 Convergence in Measure
Definition 26.7.1 (In Measure).label Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a separable metric space. For each $\eps, \delta > 0$, let
then
forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the uniform structure of convergence in measure on $\mathcal{L}^{0}(X; Y)$.
Proof. It is sufficient to check the conditions of Proposition 6.1.8:
- (FB1)
For each $\eps, \eps', \delta, \delta' > 0$,
\[U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps')\] - (UB3)
For each $\eps, \delta > 0$ and $f, g, h \in \mathcal{L}^{0}(X; Y)$,
\[\bracs{d(f, h) > \delta}\subset \bracs{d(f, g) > \delta}\cup \bracs{d(g, h) > \delta}\]so $U(\delta/2, \eps/2) \circ U(\delta/2, \eps/2) \subset U(\delta, \eps)$.
$\square$
Definition 26.7.2 (Ky Fan Metric).label Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a separable metric space, and
then:
- (1)
$\alpha$ is a metric on $L^{0}(X; Y)$.
- (2)
$\alpha$ induces the uniform structure of convergence in measure on $L^{0}(X; Y)$.
The mapping $\alpha$ is the Ky Fan metric on $L^{0}(X; Y)$.
Proof. (1): Let $f, g, h \in \mathcal{L}^{0}(X; Y)$, then
- (M)
If $\alpha(f, g) = 0$, then by continuity from above,
\[\mu\bracs{d(f, g) > 0}= \limv{n}\mu\bracs{d(f, g) > 1/n}\le \limv{n}\frac{1}{n}= 0\]so $f = g$ almost everywhere.
- (PM3)
For each $\eps > 0$,
\[\bracs{d(f, h) > \eps}\subset \bracs{d(f, g) > \eps/2}\cup \bracs{d(g, h) > \eps/2}\]so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$.
so $\alpha$ is a metric on $\mathcal{L}^{0}(X; Y)$, modulo almost everywhere equality.
(2): Let $f, g \in \mathcal{L}^{0}(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r}\le r$. Thus
On the other hand, if $\mu\bracs{d(f, g) > \eps}\le \eps$, then $d(f, g) \le \eps$. Therefore $\alpha$ induces the uniform structure of convergence in measure.$\square$
Lemma 26.7.3.label Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a separable metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_{n} \to f$ almost everywhere, then $f_{n} \to f$ in measure.
Proof. Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_{n}(x), f(x)) < \eps$ for all $n \ge N$, so
By continuity from above (Proposition 23.1.5),
$\square$
Theorem 26.7.4.label Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a Polish space, then:
- (1)
For any $\seq{f_n}\subset L^{0}(X; Y)$ that is Cauchy in measure, there exists $f \in L^{0}(X; Y)$ and a subsequence $\seq{n_k}$ such that $f_{n_k}\to f$ almost everywhere and in measure.
- (2)
$L^{0}(X; Y)$ equipped with the uniform structure of convergence in measure is complete.
Proof, [Theorem 2.30, Fol99]. (1): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
In this case, for any $K \in \natp$ and $j \ge k \ge K$,
By monotonicity and subadditivity,
so
By the First Borel-Cantelli Lemma,
Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges almost everywhere to a Borel measurable function $f \in L^{0}(X; Y)$.
Finally, for each $K \in \natp$,
so $f_{n_k}\to f$ in measure as well.
(2): Since the uniform structure of convergence in measure on $L^{0}(X; Y)$ is defined by the Ky Fan metric, completeness follows from (1) and Proposition 8.1.4.$\square$
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